Question: Simplify the following expression: $y = \dfrac{8x^2- 11x- 10}{8x + 5}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(8)}{(-10)} &=& -80 \\ {a} + {b} &=& &=& {-11} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-80$ and add them together. Remember, since $-80$ is negative, one of the factors must be negative. The factors that add up to ${-11}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${-16}$ $ \begin{eqnarray} {ab} &=& ({5})({-16}) &=& -80 \\ {a} + {b} &=& {5} + {-16} &=& -11 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({8}x^2 +{5}x) + ({-16}x {-10}) $ Factor out the common factors: $ x(8x + 5) - 2(8x + 5)$ Now factor out $(8x + 5)$ $ (8x + 5)(x - 2)$ The original expression can therefore be written: $ \dfrac{(8x + 5)(x - 2)}{8x + 5}$ We are dividing by $8x + 5$ , so $8x + 5 \neq 0$ Therefore, $x \neq -\frac{5}{8}$ This leaves us with $x - 2; x \neq -\frac{5}{8}$.